The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is . If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes [2023]

Question

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $λ_{1}$ . If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes [2023]

Smart Achievers · Accepted Answer

(1)

$From K.T.G$

$v_{RMS} = \sqrt{\frac{3 k_{B} T}{m}}$

$v_{RMS} \propto \sqrt{T} and \frac{h}{m v_{RMS}} = λ \Rightarrow λ \propto \frac{1}{\sqrt{T}}$

$\frac{λ_{2}}{λ_{1}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{300}{600}} = \frac{1}{\sqrt{2}}$

$\Rightarrow λ_{2} = \frac{λ_{1}}{\sqrt{2}}$

Solution

Q.
The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $λ_{1}$ . If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes [2023]

1 $\frac{1}{\sqrt{2}} λ_{1}$

2 $2 λ_{1}$

3 $\frac{1}{2} λ_{1}$

4 $\sqrt{2} λ_{1}$

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Solution

Q. The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes [2023]

1 12 λ1

2 2λ1

3 12 λ1

4 2 λ1